Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100;
const int INF = 1e9 + 7;
int m, n, a[N + 1][N + 1], row[N + 1], col[N + 1], srow[N + 1], scol[N + 1];
int cal(int x) {
  int res = a[1][1];
  row[1] = x;
  col[1] = a[1][1] - x;
  for (int i = 2; i <= m; ++i) {
    row[i] = a[i][1] - col[1];
    res += row[i];
    if (row[i] < 0) return -1;
  }
  for (int j = 2; j <= n; ++j) {
    col[j] = a[1][j] - row[1];
    res += col[j];
    if (col[j] < 0) return -1;
  }
  for (int i = 2; i <= m; ++i)
    for (int j = 2; j <= n; ++j)
      if (a[i][j] != row[i] + col[j]) return -1;
  return res;
}
int main() {
  cin >> m >> n;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j) cin >> a[i][j];
  int res = INF, tmp;
  for (int c = 0; c <= a[1][1]; ++c) {
    tmp = cal(c);
    if (tmp != -1 && tmp < res) {
      res = tmp;
      for (int i = 1; i <= m; ++i) srow[i] = row[i];
      for (int j = 1; j <= n; ++j) scol[j] = col[j];
    }
  }
  if (res == INF)
    cout << -1;
  else {
    cout << res << '\n';
    for (int i = 1; i <= m; ++i)
      for (int j = 0; j < srow[i]; ++j) printf("row %d\n", i);
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j < scol[i]; ++j) printf("col %d\n", i);
  }
}